\(v=a t+b t^{2}\)
\(\text { Dimensional equation is } \mathrm{LT}^{-1}\)
\(=a T=b T^{2}\)
\(\therefore \text { The dimension of } b=\frac{\mathrm{LT}^{-1}}{\mathrm{~T}^{2}}=\mathrm{LT}^{-3}\)

\(\pi=3.14 \)
\(\pi^{2} =3.14 \times 3.14 \)
\(=9.8596=9.86 \)

The number of significant figures of 6.3200 J is 5

four

Distance, D = 824.7\(\times\)10⁶ km
\(\theta=35.72"={{35.72}\over{60\times 60}}\times{{\pi}\over{180}}\) rad
\(\theta \ \rightarrow\) is angular diameter
Diameter, d = ?
\(d=D\times\theta\)
\(=824.7\times{10}^{6}\times{{35.72}\over{60\times 60}}\times{{\pi}\over{180}}\) km 
\(=824.7\times{10}^{6}\times{{35.72}\over{3600}}\times{{\pi}\over{180}}\) km
\(=824.7\times{10}^{6}\times{{35.72\times3.14}\over{3600\times180}}\)km
Diameter of Jupiter = 1.427\(\times\)10⁵ km

All units in atomic standards are more useful because they never change with time.

Length l = (5.7 ± 01.) cm
Breadth b = (3.4 ± 02.) cm
Area A with error limit = A ±\(\triangle\)A= ?
Area A = I \(\times\)b = 5.7\(\times\)3.4 = 19.38 = 19.4 cm²
\({\triangle A\over A}={\triangle l\over l}+{\triangle b\over b};\triangle A=({\triangle l\over l}+{\triangle b\over b})\)
\(\triangle A=({0.1\over 5.7}+{0.2\over 3.4})19.4=(0.0175+0.0588)\times 19.4=1.48=1.5\)
Area with error limit A = (19.4 ± 1.5) cm²

R₁ = 100 ± = 3\(\Omega\); R₂ = 150 ± 2\(\Omega\)
Equivalent resistance R =?
Equivalent resistance R = R₁+ R₂ = (100 ± 3) + (150 ± 2) = (100 + 150) ± (3 + 2)
R = (250 ± 5) \(\Omega\)

\(Mass\ m = 4\pm0.03\) kg ( m + \(\triangle\) m )
Volume \(V=5\pm.01 m^3(V+\triangle V)\)
Density =?
Error in mass \(={{\triangle m}\over{m}}={{0.03}\over{4}}\times 100\)
= 0.75%
Error in volume \(={{\triangle V}\over{V}}={{0.01}\over{5}}\times 100\)
= 0.2%
\(Density={{mass}\over{volume}}.\)
Error in density = error in mass + error in volume
= 0.75% +0.2% = 0.95%

Dimension of Pressure = [ML^-1T^-2]
Dimension of density = [ML^-3]
Dimension of Energy = [ML²T^-2]
By substituting in the given equation,
\(=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]^{-5 / 6}\left[\mathrm{ML}^{-3}\right]^{1 / 2}\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]^{1 / 3} \)
\(=\mathrm{M}^{-5 / 6+1 / 2+1 / 3} \mathrm{~L}^{5 / 6-3 / 2+2 / 3} \mathrm{~T}^{5 / 3-2 / 3}\)
=M⁰L⁰T¹=[T]

By the principle of homogeneity, a / V² is of the dimensions of pressure and b is of the dimensions of volume.
[a] = [pressure] [V²] = [ML⁻¹T⁻²] [L⁶]
= [ML⁵T^-2]
[b] = [V] = L³

(i) Solar year:
It is the time taken by the earth to complete one revolution around the sun in its orbit. 1 solar year = 365.25 average solar days.
(ii) Leap year:
The year which is divisible by 4 and in which the month of February has 29 days is called leap year.
(iii) Lunar month:
It is the time taken by the moon to complete one revolution around the earth in its orbit. 1 lunar month = 27.3 days.

(i) This system makes use of only one unit for one physical quantity, which means a rational system of units.
(ii) In this system, all the derived units can be easily obtained from basic and supplementary units, which means it is a coherent system of units.
(iii) It is a metric system which means that multiples and submultiples can be expressed as powers of 10.

n\(\propto\) I^aT^bm^c, [I] = [M^oL¹T^o]
[T] = [M¹L¹T^-2] (force)
[M] = [M¹L^-1T^o]
[M^o L^oT^-1] = [M^oL¹T^o]^a [M¹L¹T^-2]^b [M^oL^-1T^o]^C
b + c = 0
a + b - c = 0
-2b = -1 \(\Rightarrow\) b = \(1\over2\)
c =\(-{1\over2}a=1\)
n\(\propto\) \({1\over l}{\sqrt{T\over m}}\)

Let the height of the pole be h
Solution \(\frac { x+45 }{ h } \)  = cot 30^o ⇒ h =\(\frac { x+45 }{ cot\quad { 30 }^{ o } } \)
\(\frac { x }{ h } \) = cot 30^o ⇒ x = h cot 60^o
Substituting the values of x in the above equation
h = \(\frac { h\quad cot \ { 60 }^{ o }+45 }{ cot \ { 30 }^{ o } } \)
\(h \cot 30^{\circ} =h \cot 60^{\circ}+45  \)
\(h\left(\cot 30^{\circ}-\cot 60^{\circ}\right) =45  \)
\(h =\frac{45}{\cot 30^{\circ}-\cot 60^{\circ}}=\frac{45}{\sqrt{3}-\frac{1}{\sqrt{3}}}=38.97 \mathrm{~m}\)

Let G_SI be the gravitational constant in the SI system and G_cgs in the cgs system. Then 
G_SI = 6.6 10^-11 Nm² kg^-2;
G_cgs = ?
n₂ =\(n_1{ \left[ \frac { { M }_{ 1 } }{ { M }_{ 2 } } \right] }_{ }^{ a }{ \left[ \frac { L_{ 1 } }{ L_{ 2 } } \right] }_{ }^{ a }{ \left[ \frac { T_{ 1 } }{ T_{ 2 } } \right] }_{ }^{ c }\)
G_cgs = G_SI \({ \left[ \frac { { M }_{ 1 } }{ { M }_{ 2 } } \right] }_{ }^{ a }{ \left[ \frac { L_{ 1 } }{ L_{ 2 } } \right] }_{ }^{ a }{ \left[ \frac { T_{ 1 } }{ T_{ 2 } } \right] }_{ }^{ c }\)
M₁= 1 kg L₁ = 1 m T₁= 1s
M₂= 1 kg L₂ = 1 m T₂= 1s
The dimensional formula for G is M^-1L³ T^-2
a = -1 b = 3 and c =-2
G_cgs = 6.6\(\times\)10^-11 \(\left[ \frac { 1kg }{ 1g } \right] ^{ -1 }\left[ \frac { 1m }{ 1cm } \right] ^{ 3 }\left[ \frac { 1s }{ 1s } \right] ^{ -2 }\)
=  6.6\(\times\)10^-11 \(\left[ \frac { 1kg }{ { 10 }^{ -3 }kg } \right] ^{ -1 }\left[ \frac { 1m }{ { 10 }^{ -2 }m } \right] ^{ 3 }\left[ \frac { 1s }{ 1s } \right] ^{ -2 }\)
= 6.6\(\times\)10^-11\(\times\)10^-3\(\times\)10⁶\(\times\)1
G_cgs = 6.6\(\times\)10^-8 dyne cm² g^-2

\(Average \ length ={3.48+3.46+3.49+3.50+3.48\over 5}={17.41\over5}\)
= 3.482 m = 3.48 m
(Round off to 2 places of decimal point)
The absolute errors in the different measurements are
\(\triangle\) L₁= 3.48 - 3.48 = 0.00 m
\(\triangle\) L₂= 3.48 - 3.46 = 0.02 m
\(\triangle\) L₃=3.48 - 3.49 = - 0.01 m
\(\triangle\) L₄= 3.48 - 3.50 = - 0.02 m
\(\triangle\) L₅= 3.48 - 3.48 = 0.00 m .
The absolute error =\({\sum |\triangle L_i|\over 5}\)
=\(0.00+0.02+0.01+0.02+0.00\over 5\)
=\({0.05\over5}=0.01m\)
\(\therefore\) Correct length = 3.48 ± 0.01m
Percentage error =\({0.01\over 3.48}\times 100=0.29\%\)